从 SQLite 获取当前周数据和当前年份数据的总和

Question

我有一个 SQLite 数据库，sales 表如下，

| Id | quantity |     dateTime     |
------------------------------------
| 1  | 10       | 2019-12-25 12:55 |
| 2  | 05       | 2019-12-30 12:55 |
| 3  | 25       | 2020-08-23 12:55 |
| 4  | 25       | 2020-08-24 12:55 |
| 5  | 56       | 2020-08-25 12:55 |
| 6  | 25       | 2020-08-26 12:55 |
| 7  | 12       | 2020-08-27 12:55 |
| 8  | 30       | 2020-08-28 12:55 |
| 9  | 40       | 2020-08-29 12:55 |

我需要获取当前周数据（周一至周日）和当前年份数据（一月至十二月）。因此，如果我通过今天的日期，我只需要按天获取当前周销售数据组，如下所示，

如果我通过今天的日期和时间 (2020-08-28 13:55)，查询应该给我这样的当前周数据，

Day        Sold Items (SUM(quantity))
Monday     20
Tuesday    25
Wednesday  10
Thursday   50
Friday     60
Saturday   0 (If the date hasn't come yet I need to get 0)
Sunday     0

当我通过当前日期时，与当前年份数据相同，

Month      Sold Items (SUM(quantity))
JAN        20
FEB        25
MAR        10
APR        50
MAY        60
JUN        0 (If the month hasn't come yet I need to get 0)
JUL        0
...        ...

我在 SQLite 中尝试了多个查询，但无法得到我需要的。这是我尝试过的查询，

Weekly Data (This one gave me past week data also)
SELECT SUM(quantity) as  quantity, strftime('%w', dateTime) as Day
From sales
Group by strftime('%w', dateTime)

Monthly Data
SELECT SUM(quantity) as  quantity, strftime('%m', dateTime) as Month
From sales
Group by strftime('%m', dateTime)

那么任何人都可以帮助我实现这一目标吗？提前致谢。

Answer 1

你可以试试下面的 - DEMO

select day,coalesce(sum(quantity),0) as quantity
from 
(select 0 as day union all select 1 union all select 2 union all select 3 union all select 4
union all select 5 union all select 6) as d
left join sales on cast(strftime('%w', dateTime) as int)=day 
group by strftime('%w', dateTime),day
order by day

Answer 2

您可以使用以下查询来获取每周日期，我假设每个日期都有单个条目，因此不分组，否则您可以添加分组依据。

首先我们会根据输入的日期得到周历（我取的是当前日期） 然后左加入日历以获取所需的已售商品信息。

WITH seq(n) AS 
(
  SELECT 0 UNION ALL SELECT n + 1 FROM seq
  WHERE n < DATEDIFF(DAY, (SELECT DATEADD(DAY, 2 - DATEPART(WEEKDAY, GETDATE()), CAST(GETDATE() AS DATE)) [Week_Start_Date]), (Select DATEADD(DAY, 8 - DATEPART(WEEKDAY, GETDATE()), CAST(GETDATE() AS DATE)) [Week_End_Date]))
),
CALENDAR(d) AS 
(
  SELECT DATEADD(DAY, n, (SELECT DATEADD(DAY, 2 - DATEPART(WEEKDAY, GETDATE()), CAST(GETDATE() AS DATE)) [Week_Start_Date])) FROM seq
)
SELECT coalesce(QUANTITY, 0) sold_items ,DATENAME(WEEKDAY, d) week_day FROM CALENDAR a left outer join Table_WEEKDAY b
on (a.d = convert(date, b.dateTime))
ORDER BY d
OPTION (MAXRECURSION 0);

Answer 3

对于当前周的总计，您需要一个返回日期名称的 CTE 和另一个返回本周星期一的 CTE。
您必须交叉连接这些 CTE 并左连接您的表以进行聚合：

with 
  days as (
    select 1 nr, 'Monday' day union all
    select 2, 'Tuesday' union all
    select 3, 'Wednesday' union all
    select 4, 'Thursday' union all
    select 5, 'Friday' union all
    select 6, 'Saturday' union all
    select 7, 'Sunday'
  ),
  weekMonday as (
    select date(
        'now', 
        case when strftime('%w', 'now') <> '1' then '-7 day' else '0 day' end, 
        'weekday 1'
      ) monday
  )
select d.day, 
       coalesce(sum(t.quantity), 0) [Sold Items]
from days d cross join weekMonday wm
left join tablename t
on strftime('%w', t.dateTime) + 0 = d.nr % 7
and date(t.dateTime) between wm.monday and date(wm.monday, '6 day')
group by d.nr, d.day
order by d.nr

对于当年的总计，您需要一个 CTE，它返回月份名称，然后左连接表进行聚合：

with 
  months as (
    select 1 nr, 'JAN' month union all
    select 2 nr, 'FEB' union all
    select 3 nr, 'MAR' union all
    select 4 nr, 'APR' union all
    select 5 nr, 'MAY' union all
    select 6 nr, 'JUN' union all
    select 7 nr, 'JUL' union all
    select 8 nr, 'AUG' union all
    select 9 nr, 'SEP' union all
    select 10 nr, 'OCT' union all
    select 11 nr, 'NOV' union all
    select 12 nr, 'DEC'
  )
select m.month, 
       coalesce(sum(t.quantity), 0) [Sold Items]
from months m
left join tablename t
on strftime('%m', t.dateTime) + 0 = m.nr
and date(t.dateTime) between date('now','start of year') and date('now','start of year', '1 year', '-1 day')
group by m.nr, m.month
order by m.nr